Combustion and Calorific Value

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Burn one mole of methane, and you get eight hundred and ninety kilojoules of heat. Burn one mole of hydrogen, and you get only two hundred and eighty six. So methane is clearly the better fuel, right? Hmm, not so fast! Gram for gram, hydrogen beats methane by a long way. How can both be true? Today we find out. Questions on fuels and their heat do come up in JEE, so let's begin.

0:33Enthalpy of combustion: one mole, burnt completely

So, first, the enthalpy of combustion. The little triangle is the capital Greek letter delta, meaning change in, and capital H is enthalpy. The small c below it means combustion. So delta c H is the enthalpy change when one mole of a substance burns completely in oxygen. Okay? Look, there are two conditions here: exactly one mole of the substance, and complete burning, not half way. Clear? Now tell me. When something burns, heat comes out, right? So what is the sign of delta c H, positive or negative? Negative, exactly! Combustion releases heat, so delta c H is always negative. Not sometimes. Always. Got it? For example, methane. Delta c H is minus eight hundred and ninety, k J mol to the power minus one, that is, kilojoules per mole. The minus one power simply means per. With me?

1:30Writing the combustion equation

Now, how do we write a combustion equation? Look at methanol. C H three O H, a liquid, that is the l in brackets, plus three halves O two, a gas, the g in brackets, gives C O two gas, plus two H two O, liquid. The arrow means gives, and the small numbers below count the atoms. Okay? But wait. Three halves of a mole of oxygen, a fraction? Is that allowed? Yes, it is! Delta c H is for one mole of the fuel, so methanol must have a coefficient of one, and oxygen takes whatever it needs, here three halves. Right? Now look at the water. It is shown as liquid, the l, not as gas. Why? Because the standard products of combustion are C O two gas, and liquid water, at two hundred and ninety eight kelvin; the capital K on screen means kelvin. So in a standard combustion equation, water is always written as a liquid. Clear? Now suppose the water came out as steam instead. Then less heat is released: forty four kilojoules less for every eighteen grams of water. That gap is the enthalpy of condensation of water. So always write the states, and write the water as liquid. Got it?

2:49Calorific value: heat per gram

Alright, now the second big idea: calorific value. It is the heat released per gram of fuel, or per kilogram, on complete combustion. Same condition as before: the fuel must burn completely. Right? Now, you know delta c H for one mole. How do you get the heat from one gram? Divide by the molar mass, exactly! Capital M is the molar mass, the mass of one mole in grams. And the two straight bars around delta c H mean we take only its size, without the minus sign, because calorific value is heat released. Okay? Check the units. Kilojoules per mole, divided by grams per mole, that is the division sign on screen, gives k J g to the power minus one, kilojoules per gram. The moles cancel. Clear?

3:39Example 1: calorific values of methane, H2 and ethanol

Now, Example one, and it is our puzzle. Find the calorific values, in kilojoules per gram, of methane, delta c H minus eight hundred and ninety; hydrogen, minus two hundred and eighty six; and ethanol, minus one thousand three hundred and sixty seven, all in kilojoules per mole. Pause the video, and try it first. Done? Okay, let's check. Methane, C H four: one carbon, twelve, plus four hydrogens, four. So the molar mass is sixteen. Eight hundred and ninety over sixteen: fifty five point six kilojoules per gram. Okay? Now hydrogen. Only two hundred and eighty six per mole, the smallest number in the question. So hydrogen is the weakest fuel here, right? No, no, careful! That is per mole, and one mole of hydrogen, H two, weighs just two grams. Two hundred and eighty six over two: one hundred and forty three kilojoules per gram! So our puzzle is solved. Per mole, methane wins; per gram, hydrogen wins, by a long way. Clear? Ethanol, C two H five O H: two carbons, twenty four; six hydrogens, six; one oxygen, sixteen. So the molar mass is forty six. One thousand three hundred and sixty seven over forty six: twenty nine point seven kilojoules per gram. With me? So hydrogen is the perfect fuel? Not quite, there is a catch. Hydrogen has the highest calorific value per gram, but its density is very low. Okay? So a given volume of hydrogen releases very little heat, unless you compress it or liquefy it, and that is costly and dangerous. Best per gram, awkward per volume. Got it?

5:27Example 2: mass of propane to heat 5 kg of water

Now a harder one, Example two. What mass of propane, delta c H minus two thousand two hundred and twenty kilojoules per mole, must burn to heat five point zero kilograms of water from twenty to one hundred degrees Celsius, if all the heat is used? The small circle before the C means degrees Celsius. Pause the video, and try it. Done? Okay, let's check. The question says all the heat is used, so no heat is lost. First, the heat the water needs: q equals m times c times delta T, the rise in temperature. Small m is the mass of water, and small c is its specific heat, four point one eight joules per gram per kelvin. Okay? Five kilograms is five thousand grams. The rise is one hundred minus twenty, eighty degrees, and a rise of eighty degrees Celsius is a rise of eighty kelvin. So q is five thousand times four point one eight times eighty: one million, six hundred and seventy two thousand joules. Got it? Now divide by two thousand two hundred and twenty to get the moles of propane. One million, six hundred and seventy two thousand, divided by two thousand two hundred and twenty: seven hundred and fifty three moles. Right? Oh no, careful! This is the trap. The heat is in joules, but delta c H is in kilojoules. So first change q to kilojoules: one thousand six hundred and seventy two. Now small n, the number of moles, is one thousand six hundred and seventy two over two thousand two hundred and twenty: zero point seven five three moles. Clear? Propane is C three H eight: three carbons, three times twelve, plus eight hydrogens, eight times one. So the molar mass is forty four. With me? Mass equals zero point seven five three times forty four: thirty three grams. Just thirty three grams of propane heats five kilograms of water from twenty to one hundred degrees. Perfect!

7:31Quick revision: combustion enthalpy and heat per gram

So, quick revision. One: delta c H is for one mole, burnt completely in oxygen, and it is always negative. Two: the standard products are C O two gas and liquid water, at two hundred and ninety eight kelvin. Three: calorific value per gram is the size of delta c H, divided by the molar mass, and never mix up joules and kilojoules. Now try the practice problems on this page. All the best!